Period and angular frequency problem.

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A decrease in the period of a simple harmonic oscillator by 20% results in an increase in angular frequency. The relationship between period (T) and angular frequency (ω) is given by the equation T = 2π/ω. When the period is reduced to 80% of its original value, the new angular frequency can be calculated as ω = 2π/T * 1.25. This indicates that the angular frequency increases by 25%. Thus, a 20% reduction in period leads to a 25% increase in angular frequency.
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Homework Statement



If period of simple harmonic graph of electric current goes down by 20%, how will change the angular frequency?
Will it increase by 25π%? 25%? 50%?...

Homework Equations



T = 2π/ω


The Attempt at a Solution



So by going down by 20%, so 100%-20%, I have 80%*T=2π/ω. 0.8*T=2π/ω. 8T/10=2π/ω. 4T/5=2π/ω. ω=2π/T *(5/4) = 2π/T * 1.25
So by this, angular frequency increases by 25%?
 
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